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把数字拆分成2的幂的和

问题:任何数都能分解成2的幂,比如71+1+1+1+1+1+11+1+1+1+1+21+1+1+2+21+2+2+2

问题:

    任何数都能分解成2的幂,比如

  7=1+1+1+1+1+1+1

    =1+1+1+1+1+2

    =1+1+1+2+2

    =1+2+2+2

    =1+1+1+4

    =1+2+4

共有6种分解方式,设f(n)为任意正整数可能分解总数,比如f(7)=6

写个算法,输入数,求出其分解的总数。

 

思路:

    先按照树形结构,把一个数可能的2的幂的子数记录下来,比如7拆分成7个1,3个2,1个4。从高到底遍历所有可能的搭配。


import math  
import copy

def get_distribute_for_number(number):
distribute = {}
distribute[0] = number
for i in range(0, int(math.log(number, 2) + 1)):
if i not in distribute:
break
count_i = distribute[i]
while count_i >= 2:
count_i -= 2
if i + 1 not in distribute:
distribute[i + 1] = 0
distribute[i + 1] += 1
return distribute

def count_by_distribute(distribute, number, parent_expr):
if number == 0:
print("expr : %s"%parent_expr[:-3])
return 1
max_leaf = len(distribute) - 1
if max_leaf == 0:
print("expr : %s"%(parent_expr + " 1 * %d"%number))
return 1
curr_distribute = copy.copy(distribute)
max_leaf_value = 2 ** max_leaf
max_leaf_num = curr_distribute.pop(max_leaf)
count = 0
for i in range(0, max_leaf_num + 1):
left = number - max_leaf_value * i
expr = parent_expr
expr += "%d * %d"%(max_leaf_value, i)
expr += " + "
if left <0:
break
count_left = count_by_distribute(curr_distribute, left, expr)
count += count_left
#print("current distribute is ")
#print(curr_distribute)
#print("kept %d leaves of %d"%(i, max_leaf_value))
#print("distributing num for %d is %d"%(left, count_left))
return count

number = input("Input Number:")
distribute = get_distribute_for_number(number)
print("Distribute for num is")
print(distribute)
count = count_by_distribute(distribute, number, "")
print("Number %d can be distributed by %d ways"%(number, count))



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李辉国0321_817
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